Problem: Simplify the following expression and state the condition under which the simplification is valid. $q = \dfrac{p^3 - 7p^2 - 8p}{-8p^2 + 80p - 128}$
Solution: First factor out the greatest common factors in the numerator and in the denominator. $ q = \dfrac {p(p^2 - 7p - 8)} {-8(p^2 - 10p + 16)} $ $ q = -\dfrac{p}{8} \cdot \dfrac{p^2 - 7p - 8}{p^2 - 10p + 16} $ Next factor the numerator and denominator. $ q = - \dfrac{p}{8} \cdot \dfrac{(p - 8)(p + 1)}{(p - 8)(p - 2)}$ Assuming $p \neq 8$ , we can cancel the $p - 8$ $ q = - \dfrac{p}{8} \cdot \dfrac{p + 1}{p - 2}$ Therefore: $ q = \dfrac{ -p(p + 1)}{ 8(p - 2)}$, $p \neq 8$